Why is row $n = 2^x$ in Pascal's triangle have all even numbers except the $1$'s? In row $n = 2^x$, $x$ being a positive integer, in the Pascal's triangle, all entries except the two $1$'s in extreme left and right are even.
I tried to prove but I couldn't.
Here is my try:-
Every entry is of the form $\frac{(2^x)!}{(k!)([2^x]-k)!}$
I counted the no.of 2's in the prime factorisation of $(2^x)!$ in the following way:-
No.of multiples of $2 = \frac{2^x}2 = 2^{x-1}$
No.of multiples of $4 = \frac{2^x}{2^2}= 2^{x-2}$
Similarily upto no.of multiples of $2^x = 1$.
So total no.of 2's in prime factorisation is
$$
2^{x-1} + 2^{x-2} + \cdots + 2 + 1 = 2^x-1
$$
But I cannot prove that the no.of 2's in the denominator of each entry will be less than $2^{x-1}$.
Can I get some hints/help?
Thank you.
 A: Here's a simple solution,
$$(1+x)^2 \equiv 1+x^2 \mod 2$$
$$\rightarrow (1+x)^{2^2} \equiv (1+x^2)^2 \equiv 1+x^4 \mod 2$$
if you keep going,
$$(1+x)^{2^n}\equiv 1+x^{2^n}\mod 2$$
implying that all the binomials of the form $\binom{2^n}{k}\equiv 0 \mod 2\qquad \forall 0<k<n$ 

For more in-depth analysis I suggest you have a look at Lucas's theorem
The theorem says that for any prime $p$ and numbers $m,n$ which can be written as,
$$m = m_k p^k + m_{k-1}p^{k-1}+ \dots+m_0$$
$$n = n_k p^k + n_{k-1}p^{k-1}+ \dots+n_0$$
we have,
$$\binom{m}{n} = \binom{m_k}{n_k}\binom{m_{k-1}}{n_{k-1}}\dots \binom{m_{0}}{n_{0}} \mod p$$
$2^x$ in binary representation is $10...0$ ($x$ zeroes).
Any number $0<k<2^n$ will have binary representation $0b_1b_2...b_x$ where $b_i\in (0,1)$ but not all are 0.
$$\binom{2^x}{k}\equiv \binom{1}{0}\prod_{i=1}^{i=x}\binom{0}{b_i}\mod 2$$
We have $\binom{m}{n}=0$ for $m<n$. Since all the $b_i$ cannot be zero simutaneously ($k\neq 0$ or $2^x$) we have,
$$\binom{2^x}{k}\equiv 0 \mod 2 \qquad 0<k<2^x$$
A: The key observation is that if $1\le k < 2^x$, then $k$ and $2^x-k$ contain exactly the same power of $2$. So when we write
$$\binom{2^x}{k}=\frac{2^x(2^x-1)\ldots(2^x-k+1)}{k(k-1)\ldots2\cdot 1}$$
we can cancel the powers of $2$ in $(2^x-r)$ in the numerator with the powers of $2$ in $r$ in the denominator, leaving us with just the powers of $2$ contained in
$$\frac{2^x}{k}$$
which is obviously even if $1\le k < 2^x$.
A: Suppose we are looking at the binomial coefficient $$\binom {2^n}m=\frac {(2^n)!}{m!(2^n-m)!}=\frac {2^n}m\cdot\frac {2^n-1}1\cdot\frac{2^n-2}2\dots \cdot \frac {2^n-(m-1)}{m-1}$$
(assuming $m\gt 1$: we have only the first term for $m=1$, and for $m=0$ the empty product or straightforwardly the value $1$)
Now suppose $2^n\gt r \gt 0$ and that $r=2^st$ where $t$ is odd, and we have $s\lt n$, then all the fractions in the product apart from the first are of the form $$\frac {2^n-r}r=\frac {2^n-2^st}{2^st}=\frac {2^{n-s}-t}{t}$$and this is a fraction with odd numerator and denominator. On the other hand with $\frac {2^n}m$ we have that $m$ is divisible by a lower power of $2$ than $2^n$ unless $m=2^n$. So we have a positive power of $2$ in the product.
This can be adapted for powers of any prime.
A: We'll write the greatest integer $\le y$ as $\lfloor y\rfloor$, and define $\{y\}:=y-\lfloor y\rfloor$. The greatest integer $m$ for which $2^m|F!$ is $\sum_{j\ge 0}\left\lfloor\frac{F}{2^j}\right\rfloor$. (This sum's nonzero terms are those with $j\le\lfloor\log_2F\rfloor$.) Thus the greatest integer $m$ for which $2^m|\binom{2^x}{k}$ is $$M:=\sum_{j=0}^x\left(\left\lfloor\frac{2^x}{2^j}\right\rfloor-\left\lfloor\frac{k}{2^j}\right\rfloor-\left\lfloor\frac{2^x-k}{2^j}\right\rfloor\right)=\sum_{j=0}^x\left(\left\{\frac{k}{2^j}\right\}+\left\{\frac{2^x-k}{2^j}\right\}\right).$$This is positive if $1\le k\le2^x-1$.
A: First note that in row 2 of Pascal's triangle is $1,2,1$, so the statement is true for $n=2$. Now let $x$ be arbitrarily chosen and suppose the statement holds for $n=2^{x}$.
In particular this means that for 
$$(1+x)^{2^{x}}=a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n-1}x^{n-1}+a_{n}x^{n}$$
we have $a_{1},...,a_{n-1}$ are all even numbers and $a_{0}=a_{n}=1$. So
$$(1+x)^{2^{x+1}}=b_{0}+b_{1}x+b_{2}x^{2}+...+b_{2n-1}x^{2n-1}+b_{2n}x^{2n}$$
$$=(1+a_{1}x+a_{2}x^{2}+...+a_{n-1}x^{n-1}+x^{n})^{2}$$
$$=1+x^{2n}+\sum^{n-1}_{i=1}a_{i}^{2}x^{2i}+\sum_{0\leq i<j\leq n}2a_{i}a_{j}x^{i+j}$$
Since for $0\leq i<j\leq n$ we find that $0<i+j<2n$ we find that $b_{0}=b_{2n}=1$ and for $0<k<2n$ we find that $b_{k}$ is the sum of even numbers and therefore even.
